Basic power rule ((a^b)^c = a^(b*c)) does not work
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Just cracked down a large problem in my code to a strange phenomenon. A very simple power rule doesn't seem to work here.
b = 5i; %complex, this should have something to do with it
c = 0.1; %for intergers it seems to work
x1 = exp(b)^c;
x2 = exp(b*c);
x1 =
0.9918 - 0.1280i
x2 =
0.8776 + 0.4794i
Shouldn't they be equal?
I just do not unterstand..
댓글 수: 2
Greg Dionne
2019년 2월 1일
편집: Greg Dionne
2019년 2월 1일
I can't tell if this is homework or not, so here's a thought experiment:
At what value of x does the following print different values for your two expressions?
for x=0.01:.01:5;
y = 0.5/x; % make sure x*y = 0.5
expression1 = exp(1i*x)^y;
expression2 = exp(1i*x*y);
fprintf('%f: %f + %fi ?= %f + %fi\n',x, ...
real(expression1), imag(expression1), ...
real(expression2), imag(expression2));
end
채택된 답변
Luna
2019년 2월 1일
편집: Luna
2019년 2월 1일
By the definition Euler's equation: z = x + iy means
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202375/image.png)
So the two things might be different:
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202376/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202377/image.png)
x2 = exp(b*c) is equal to exp(c)^b actually.
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202378/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202376/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202385/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202377/image.png)
![](https://www.mathworks.com/matlabcentral/answers/uploaded_files/202382/image.png)
But those two above are not equal.
댓글 수: 2
Walter Roberson
2019년 2월 1일
(I notice the latex is cut off at the top of each section. I have reported this to Mathworks a second ago.)
추가 답변 (1개)
James Tursa
2019년 2월 1일
편집: James Tursa
2019년 2월 1일
This has been discussed in this forum before. Raising complex numbers to a power is a multi-valued operation. MATLAB picks one of those results, which may or may not agree with the result you get by rearranging the operations. E.g.,
>> b = 5i;
>> c = 0.1;
>> exp(b)
ans =
0.283662185463226 - 0.958924274663138i
>> exp(b+2*pi*i) % You can add 2*pi to get same result
ans =
0.283662185463226 - 0.958924274663139i
>> exp(b+4*pi*i) % You can add 4*pi to get same result
ans =
0.283662185463226 - 0.958924274663139i
>> exp(b+18*pi*i) % In general, you can add k*2*pi
ans =
0.283662185463224 - 0.958924274663139i
>>
>> exp(b)^c
ans =
0.991778467700342 - 0.127966679280045i % The MATLAB answer for this expression
>> exp(b*c)
ans =
0.877582561890373 + 0.479425538604203i % This particular rearrangement doesn't match
>> exp((b+18*pi*i)*c)
ans =
0.991778467700342 - 0.127966679280045i % But this one does
You can probably look at the doc to see the rules for how MATLAB picks the result for these mutli-valued issues. But the bottom line is if you have code that depends on multi-valued calculations like this, you need to account for it in your code logic. MATLAB isn't going to know which one you want picked, and you shouldn't expect it to.
댓글 수: 2
Stephen23
2019년 2월 2일
"18pi seems a bit arbitrary"
It is entirely arbitrary. That is exactly the point.
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